Let \( y = y(x) \) be the solution of the differential equation \[\left( 1 + x^2 \right) \frac{dy}{dx} + y = e^{\tan^{-1}x}, \, y(1) = 0.\]Then \( y(0) \) is:
- \( \frac{1}{4} \left( e^{\pi/2} - 1 \right) \)
- \( \frac{1}{2} \left( 1 - e^{\pi/2} \right) \)
- \( \frac{1}{4} \left( 1 - e^{\pi/2} \right) \)
- \( \frac{1}{2} \left( e^{\pi/2} - 1 \right) \)
The Correct Option is B
Approach Solution - 1
To solve the given differential equation, we notice that it is a first-order linear differential equation of the form:
\[(1 + x^2) \frac{dy}{dx} + y = e^{\tan^{-1}x}.\]We need to make this equation into the standard form:
\[\frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{e^{\tan^{-1}x}}{1 + x^2}.\]The integrating factor (IF) for this equation is given by:
\[IF = e^{\int \frac{1}{1+x^2} \, dx}= e^{\tan^{-1}x}.\]Multiplying the entire differential equation by this integrating factor, we have:
\[e^{\tan^{-1}x}\left(\frac{dy}{dx} + \frac{y}{1 + x^2}\right) = \frac{e^{2\tan^{-1}x}}{1 + x^2}.\]This simplifies to:
\[\frac{d}{dx}\left(ye^{\tan^{-1}x}\right) = \frac{e^{2\tan^{-1}x}}{1 + x^2}.\]Integrating both sides, we have:
\[ye^{\tan^{-1}x} = \int \frac{e^{2\tan^{-1}x}}{1 + x^2} \, dx + C.\]Now, to evaluate the integral, let \(t = \tan^{-1}x\), which implies \(\frac{dt}{dx} = \frac{1}{1+x^2}\) or \(dx = (1 + x^2) \, dt\). Thus, the integral becomes:
\[\int e^{2t} \, dt = \frac{e^{2t}}{2} + C.\]Substituting back, we have:
\[ye^{\tan^{-1}x} = \frac{e^{2\tan^{-1}x}}{2} + C.\]Substituting the initial condition \(y(1) = 0\) when \(x = 1\):
\[0 = \frac{e^{\pi/4}}{2} + C.\]This gives us:
\[C = -\frac{e^{\pi/2}}{2}.\]Substituting C back into the solution, we have:
\[ye^{\tan^{-1}x} = \frac{e^{2\tan^{-1}x}}{2} - \frac{e^{\pi/2}}{2}.\]Therefore, solving for \(y\), we have:
\[y = \frac{e^{\tan^{-1}x}}{2} - \frac{e^{\tan^{-1}x-\pi/2}}{2}.\]To find \(y(0)\), substitute \(x = 0\), which gives:
\[y(0) = \frac{1}{2} - \frac{1}{2} e^{\pi/2}.\]Thus, the value of \(y(0)\) is:
\( \frac{1}{2} \left( 1 - e^{\pi/2} \right) \)
Approach Solution -2
The given differential equation is:
\[ \frac{dy}{dx} + \frac{y}{1+x^2} = \frac{e^{\tan^{-1}x}}{1+x^2}. \]
Integrating factor (I.F.):
\[ \text{I.F.} = e^{\int \frac{1}{1+x^2} dx} = e^{\tan^{-1}x}. \]
Multiply through by I.F.:
\[ y \cdot e^{\tan^{-1}x} = \int e^{\tan^{-1}x} \cdot \frac{e^{\tan^{-1}x}}{1+x^2} dx + C. \]
Simplify:
\[ y \cdot e^{\tan^{-1}x} = \int e^{2\tan^{-1}x} \cdot \frac{1}{1+x^2} dx + C. \]
Substitute:
\[ \tan^{-1}x = z, \quad \frac{1}{1+x^2} dx = dz. \]
Thus:
\[ y \cdot e^{\tan^{-1}x} = \frac{e^{2z}}{2} + C. \]
Apply initial condition: \( y(1) = 0 \), \( \tan^{-1}(1) = \frac{\pi}{4} \):
\[ 0 = \frac{e^{\pi/2}}{2} + C \implies C = -\frac{e^{\pi/2}}{2}. \]
Finally:
\[ y \cdot e^{\tan^{-1}x} = \frac{e^{2\tan^{-1}x}}{2} - \frac{e^{\pi/2}}{2}. \]
At \( x = 0 \), \( \tan^{-1}(0) = 0 \):
\[ y(0) = \frac{1}{2} \left( 1 - e^{-\pi/2} \right). \]
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